All vertex pairs connected with exactly k edges in a graph

Given a directed graph represented as an adjacency matrix and an integer ‘k’, the task is to find all the vertex pairs that are connected with exactly ‘k’ edges.
Also, find the number of ways in which the two vertices can be linked in exactly k edges.

Examples :

Input : k = 3 and graph :
0 1 0 0 0 
0 0 1 0 0 
0 0 0 1 1 
1 0 0 0 0 
0 0 1 0 0 
Output :
1 -> 4 in 1 way(s)
1 -> 5 in 1 way(s)
2 -> 1 in 1 way(s)
2 -> 3 in 1 way(s)
3 -> 2 in 1 way(s)
3 -> 4 in 1 way(s)
3 -> 5 in 1 way(s)
4 -> 3 in 1 way(s)
5 -> 1 in 1 way(s)
5 -> 3 in 1 way(s)

Input : k = 2 and graph :
0 0 0 
1 0 1 
0 1 0 
Output :
2 -> 2 in 1 way(s)
3 -> 1 in 1 way(s)
3 -> 3 in 1 way(s)

Approach :

  • We will multiply the adjacency matrix with itself ‘k’ number of times.
  • In the resultant matrix, res[i][j] will be the number of ways in which vertex ‘j’ can be reached from vertex ‘i’ covering exactly ‘k’ edges.

Below is the implementation of the above approach :

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// Java implementation of the approach
public class KPaths {
  
    // Function to multiply two square matrices
    static int[][] multiplyMatrices(int[][] arr1, int[][] arr2)
    {
        int order = arr1.length;
        int[][] ans = new int[order][order];
        for (int i = 0; i < order; i++) {
            for (int j = 0; j < order; j++) {
                for (int k = 0; k < order; k++) {
                    ans[i][j] += arr1[i][k] * arr2[k][j];
                }
            }
        }
        return ans;
    }
  
    // Function to find all the pairs that
    // can be connected with exactly 'k' edges
    static void solve(int[][] arr, int k)
    {
        int[][] res = new int[arr.length][arr[0].length];
  
        // copying arr to res,
        // which is the result for k=1
        for (int i = 0; i < res.length; i++)
            for (int j = 0; j < res.length; j++)
                res[i][j] = arr[i][j];
  
        // multiplying arr with itself
        // the required number of times
        for (int i = 2; i <= k; i++)
            res = multiplyMatrices(res, arr);
  
        for (int i = 0; i < res.length; i++)
            for (int j = 0; j < res.length; j++)
  
                // if there is a path between 'i'
                // and 'j' in exactly 'k' edges
                if (res[i][j] > 0)
                    System.out.println(i + " -> " + j + 
                        " in " + res[i][j] + " way(s)");
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int[][] arr = new int[5][5];
        arr[0][1] = 1;
        arr[1][2] = 1;
        arr[2][3] = 1;
        arr[2][4] = 1;
        arr[3][0] = 1;
        arr[4][2] = 1;
        int k = 3;
        solve(arr, k);
    }
}

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Output:

0 -> 3 in 1 way(s)
0 -> 4 in 1 way(s)
1 -> 0 in 1 way(s)
1 -> 2 in 1 way(s)
2 -> 1 in 1 way(s)
2 -> 3 in 1 way(s)
2 -> 4 in 1 way(s)
3 -> 2 in 1 way(s)
4 -> 0 in 1 way(s)
4 -> 2 in 1 way(s)


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